Question Number 27684 by abdo imad last updated on 12/Jan/18
![1) prove the existence of the integral I=∫_0 ^(π/2) ((ln(1+cosx))/(cosx))dx 2)prove that I= ∫∫_D ((siny)/(1+cosx cosy))dxdy with D=[0,(π/2)]^2 3)find the value of I.](https://www.tinkutara.com/question/Q27684.png)
$$\left.\mathrm{1}\right)\:{prove}\:{the}\:{existence}\:{of}\:{the}\:{integral} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{ln}\left(\mathrm{1}+{cosx}\right)}{{cosx}}{dx} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:{I}=\:\int\int_{{D}} \:\:\frac{{siny}}{\mathrm{1}+{cosx}\:{cosy}}{dxdy}\:{with}\: \\ $$$${D}=\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{I}. \\ $$
Commented by abdo imad last updated on 16/Jan/18
![1) the convergence of I is easy lim_(x→(π/2)) ((ln(1+cosx))/(cosx)) = lim_(x→(π/2)) ((−sinx)/(−sinx(1+cosx)))= 1 (by hospital theorem ) so the function is local integrable in [0,(π/2)[ 2) I= ∫∫_(0≤x≤ (π/2) and 0≤y≤ (π/2)) ((siny)/(1+cosx cosy)) dx dy I= ∫_0 ^(π/2) ( ∫_0 ^(π/2) (1/(cosx))( −( ((1+cosx cosy)^, )/(1+cosx cosy)))dy)dx I= ∫_0 ^(π/2) (1/(cosx)) [−ln/1+cosx cosy/]_(y=0) ^(y=(π/2)) ]dx I=∫_0 ^(π/(2 )) ((ln(1+cosx))/(cosx))dx 3) I=∫_0 ^(π/2) (∫_0 ^(π/2) (dx/(1+cosy cosx)))siny dy but J= ∫_0 ^(π/2) (dx/(1+cosy cosx))= ∫_0 ^(π/2) (dx/(1+λ cosx)) (λ=cosy)and the ch. tan((x/2))=t give J= ∫_0^ ^1 (((2dt)/(1+t^2 ))/(1+λ((1−t^2 )/(1+t^2 ))))= ∫_0 ^1 ((2dt)/(1+t^2 +λ−λt^2 )) = ∫_0 ^1 ((2dt)/((1−λ)t^2 +1+λ))= (2/(1−λ)) ∫_0 ^1 (dt/(t^2 +((√(((1+λ)/(1−λ)))))))2 = (2/( (√(1−λ^2 )))) ∫_0 ^(√((1−λ)/(1+λ))) (du/(1+u^2 )) (ch.t=(√((1+λ)/(1−λ))) u) =(2/( (√(1−λ^2 )))) artan( (√((1−λ)/(1+λ)))))= (2/(siny)) artan(tan((y/(2)))))= (y/(siny)) I= ∫_0 ^(π/2) (y/(siny)) sinydy= [(1/2) y^2 ]_0 ^(π/2) = (1/2) (π^2 /4)= (π^2 /8) .](https://www.tinkutara.com/question/Q27916.png)
$$\left.\mathrm{1}\right)\:{the}\:{convergence}\:{of}\:{I}\:{is}\:{easy}\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\frac{{ln}\left(\mathrm{1}+{cosx}\right)}{{cosx}} \\ $$$$=\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\:\frac{−{sinx}}{−{sinx}\left(\mathrm{1}+{cosx}\right)}=\:\mathrm{1}\:\:\left({by}\:{hospital}\:{theorem}\:\:\right) \\ $$$${so}\:{the}\:{function}\:{is}\:{local}\:{integrable}\:{in}\:\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\right.\right. \\ $$$$\left.\mathrm{2}\right)\:{I}=\:\int\int_{\mathrm{0}\leqslant{x}\leqslant\:\frac{\pi}{\mathrm{2}}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\:\frac{\pi}{\mathrm{2}}} \:\:\frac{{siny}}{\mathrm{1}+{cosx}\:{cosy}}\:{dx}\:{dy} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\mathrm{1}}{{cosx}}\left(\:\:−\left(\:\frac{\left.\mathrm{1}+{cosx}\:{cosy}\right)^{,} }{\mathrm{1}+{cosx}\:{cosy}}\right){dy}\right){dx}\right. \\ $$$$\left.{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{{cosx}}\:\left[−{ln}/\mathrm{1}+{cosx}\:{cosy}/\right]_{{y}=\mathrm{0}} ^{{y}=\frac{\pi}{\mathrm{2}}} \right]{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} \:\:\:\:\frac{{ln}\left(\mathrm{1}+{cosx}\right)}{{cosx}}{dx} \\ $$$$\left.\mathrm{3}\right)\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{cosy}\:{cosx}}\right){siny}\:{dy}\:\:{but} \\ $$$$\:{J}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{cosy}\:{cosx}}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+\lambda\:{cosx}}\:\left(\lambda={cosy}\right){and}\:{the}\:{ch}. \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\:{give}\:\:{J}=\:\int_{\mathrm{0}^{} } ^{\mathrm{1}} \:\:\:\:\:\frac{\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\lambda\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\lambda−\lambda{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}−\lambda\right){t}^{\mathrm{2}} \:+\mathrm{1}+\lambda}=\:\frac{\mathrm{2}}{\mathrm{1}−\lambda}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} +\left(\sqrt{\left.\frac{\mathrm{1}+\lambda}{\mathrm{1}−\lambda}\right)}\right.}\mathrm{2} \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}\:\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}}} \:\:\:\:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:\:\:\:\:\left({ch}.{t}=\sqrt{\frac{\mathrm{1}+\lambda}{\mathrm{1}−\lambda}}\:\:{u}\right) \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}\:{artan}\left(\:\sqrt{\frac{\mathrm{1}−\lambda}{\left.\mathrm{1}+\lambda\right)}}\right)=\:\frac{\mathrm{2}}{{siny}}\:{artan}\left({tan}\left(\frac{{y}}{\left.\mathrm{2}\right)}\right)\right)=\:\frac{{y}}{{siny}} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{y}}{{siny}}\:{sinydy}=\:\left[\frac{\mathrm{1}}{\mathrm{2}}\:{y}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$